Help with Cauchy Integral Formula

[Paradoxx]

Use Cauchy Integral Formula to solve:

∫ [(5z² - 3z + 2)/(z-1)³] dz

C is any closed simple curve involving z=1. (z is a complex)

Thanks and sorry for my poor english, it's not my first language.

 

[Dick]

Use Cauchy Integral Formula to solve:

∫ [(5z² - 3z + 2)/(z-1)³] dz

C is any closed simple curve involving z=1. (z is a complex)

Thanks and sorry for my poor english, it's not my first language.

State the Cauchy Integral Formula. Then express your function in a simpler form using partial fractions. Do something.

 

[Paradoxx]

Ok, I put it in partial fractions:

[(5z² - 3z + 2)/(z-1)³] = 5/(z-1) + 7/(z-1)² + 4/(z-1)³

But I have no idea how to solve it using Cauchy... can anybody at least give me a hint? Or try to explain something, a next step?
I've seen a few videos and examples around internet but when I try to solve it, I just can't apply... I really need help on this.

 

[HallsofIvy]

You were asked to "state the Cauchy Integral Formula". Are you saying you do not know what it is? Surely if your textbook refers to it, it must be stated in that chapter.

(You really don't need the full "integral formula". It is sufficient to know that the integral around two different paths is the same as long as there are no singularities in between the two paths. Here you can take the integral around a circle centered at 0, with radius 1.)

 

[Paradoxx]

Like, with the partial fractions I will have now

i2π(f(5)) + (f(7)) + (f(4))

where the f(a) = (5z² - 3z + 2).


Is this that I have to do?

 

[vanhees71]

State the integral formula! Then you should know, what's to do. Of course you must not take any curve going through the singularity at $z=1$ as suggested by HallsofIvy, but any closed curve with this point in the interior of the area enclosed by it!

 

[Paradoxx]

Sorry, I know I'm really dumb ...but anyways

I have this Cauchy Int. Formula : ∫F(z)/((z-z₀))ⁿ

My doubt is about this countour, it just says it's a simple closed one with z =1.In this case, do I have a circle(for exemple) centred in (0,0) with r=1?

Then my z₀=1
and my n= 3
and my f(z)= 5z²-3z+2

Then I have that Cauchy In. Formula = [2πi /(3)!] f⁽³⁾(z⁰)

And f⁽³⁾ means that I have to derivate f(z) 3 time, is it?

 

[Dick]

Sorry, I know I'm really dumb ...but anyways

I have this Cauchy Int. Formula : ∫F(z)/((z-z₀))ⁿ

My doubt is about this countour, it just says it's a simple closed one with z =1.In this case, do I have a circle(for exemple) centred in (0,0) with r=1?

Then my z₀=1
and my n= 3
and my f(z)= 5z²-3z+2

Then I have that Cauchy In. Formula = [2πi /(3)!] f⁽³⁾(z⁰)

And f⁽³⁾ means that I have to derivate f(z) 3 time, is it?

You want a circle around the point z=1. That's the point (1,0) if you are looking at x,y coordinates in the complex plane. And the radius or shape doesn't really matter. It just has to wind around z=1. Apply the Cauchy Integral Formula to each term in your partial fractions form. Or if you can do it all at once like you are doing, but read the Formula statement more carefully if you think you differentiate three times. You don't.

 

[Paradoxx]

Here's the formula I know for cauchy's integral

Where n is the radius, and f³(z) means I have to derivate F(z) 3times. At least it's what's on the only exercise we had in class...(f(z) was an exponencial at tha case)

Decomposing it in partial fraction doesn't help. If was a function type of 1/((z-a)(z-b)) or a quadratic function I would know...then I would appy another version of this formula.

But 1/(z-1)³ gives me fractions with power on the denominator, it doesn't help.

I have this specific function(that I posted) when you have powers on the denominator, but it demands de radius...


I'm stucked, someone explain how I must do it?!

 

[Dick]

Here's the formula I know for cauchy's integral

Where n is the radius, and f³(z) means I have to derivate F(z) 3times. At least it's what's on the only exercise we had in class...(f(z) was an exponencial at tha case)

Decomposing it in partial fraction doesn't help. If was a function type of 1/((z-a)(z-b)) or a quadratic function I would know...then I would appy another version of this formula.

But 1/(z-1)³ gives me fractions with power on the denominator, it doesn't help.

I have this specific function(that I posted) when you have powers on the denominator, but it demands de radius...I'm stucked, someone explain how I must do it?!

I can barely read the formula you've attached but it looks like you have $(z-z_0)^{n+1}$ in the denominator. So if you have $(z-1)^3$ then n+1=3 and you want to use n=2 in your formula. Not n=3. Why are you saying n is the radius? I don't see any radius in that formula.

 

[Paradoxx]

If you click with the right button and open it in a new tab you'll see it clearly. Or click in the image, then will appear a black image when you can barely see it, just click in this black image and it will open a new tab with the formula.
Anyaway is just http://upload.wikimedia.org/math/2/1/1/211c06a5ce0bfe247bbe60d43dddfea1.png

n is the |z|, so n is the radius. In the example in class, that's the information given...

In the problem I wrote C is any closed simple curve involving z=1. Then I assumed |z| = 1, where z = x+ iy
so x² + y² = 1²

I thought it wrong?

 

[Dick]

If you click with the right button and open it in a new tab you'll see it clearly. Or click in the image, then will appear a black image when you can barely see it, just click in this black image and it will open a new tab with the formula.
Anyaway is just http://upload.wikimedia.org/math/2/1/1/211c06a5ce0bfe247bbe60d43dddfea1.png

n is the |z|, so n is the radius. In the example in class, that's the information given...

In the problem I wrote C is any closed simple curve involving z=1. Then I assumed |z| = 1, where z = x+ iy
so x² + y² = 1²

I thought it wrong?

Thanks, that link works better. Your information is wrong. n isn't the radius. n+1 is supposed to be the power in the denominator. You don't need to know the radius. You just need to know the curve circles z=1. And x^2+y^2=1 doesn't do that. It goes through z=1. A better curve is (x-1)^2+y^2=1, that's actually a circle around z=1. z=1 is the center.

 

[Paradoxx]

So n+1=3, n=2

And according to that formula I have to do f''(z₀)

Where my F(z) = 5z²-3z+2
F''(z)=10

F''(z)[(2∏i)/2!] = ∫5z²-3z+2 / (x-1)³ = 10i∏

Is it?

 

[Dick]

So n+1=3, n=2

And according to that formula I have to do f''(z₀)

Where my F(z) = 5z²-3z+2
F''(z)=10

F''(z)[(2∏i)/2!] = ∫5z²-3z+2 / (x-1)³ = 10i∏

Is it?

Now you've got it.

 

[Paradoxx]

Yay! Thanks...I've got to do lots of exercises now to practice.

 

[Dick]

Yay! Thanks...I've got to do lots of exercises now to practice.

Great. One last thing to watch out for is that that formula also assumes the integration is going counterclockwise around z=1. If it were clockwise you'd have to reverse the sign.